Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
struct P {
  int x, y;
  P(int x, int y) : x(x), y(y) {}
};
P operator-(const P &a, const P &b) { return P(a.x - b.x, a.y - b.y); }
long long operator^(const P &a, const P &b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
int ay, alx, arx;
int n;
int bl[MAXN], br[MAXN];
int nq;
int qx[MAXN], qy[MAXN];
int bsum[MAXN + 1];
int calcside(int ax, int ay, int bx, int by, int cx, int cy) {
  P a(ax, ay), b(bx, by), c(cx, cy);
  long long res = (c - a) ^ (b - a);
  return res < 0 ? -1 : res > 0 ? +1 : 0;
}
double calc(int i) {
  int l = -1, r = n;
  while (l + 1 < r) {
    int m = l + (r - l) / 2;
    int side = calcside(alx, ay, qx[i], qy[i], br[m], 0);
    if (side >= 0)
      r = m;
    else
      l = m;
  }
  int fst = r;
  l = -1, r = n;
  while (l + 1 < r) {
    int m = l + (r - l) / 2;
    int side = calcside(arx, ay, qx[i], qy[i], bl[m], 0);
    if (side <= 0)
      l = m;
    else
      r = m;
  }
  int lst = l;
  if (fst > lst) return 0;
  bool fstpart = calcside(alx, ay, qx[i], qy[i], bl[fst], 0) < 0;
  bool lstpart = calcside(arx, ay, qx[i], qy[i], br[lst], 0) > 0;
  if (fst == lst && fstpart && lstpart) return arx - alx;
  double ret = 0;
  if (fstpart) {
    double x = qx[i] + 1.0 * (qy[i] - ay) * (br[fst] - qx[i]) / qy[i];
    ret += x - alx;
    ++fst;
  }
  if (lstpart) {
    double x = qx[i] + 1.0 * (qy[i] - ay) * (bl[lst] - qx[i]) / qy[i];
    ret += arx - x;
    --lst;
  }
  int mid = bsum[lst + 1] - bsum[fst];
  ret += 1.0 * mid * (qy[i] - ay) / qy[i];
  return ret;
}
void run() {
  scanf("%d%d%d", &ay, &alx, &arx);
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &bl[i], &br[i]);
  scanf("%d", &nq);
  for (int i = (0); i < (nq); ++i) scanf("%d%d", &qx[i], &qy[i]);
  bsum[0] = 0;
  for (int i = (0); i < (n); ++i) bsum[i + 1] = bsum[i] + br[i] - bl[i];
  for (int i = (0); i < (nq); ++i) {
    double ans = calc(i);
    printf("%.15lf\n", ans);
  }
}
int main() {
  run();
  return 0;
}